Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $t \neq 0$. $p = \dfrac{t^2 - 11t + 24}{3t^2 + 12t} \div \dfrac{t - 3}{-6t - 24} $
Solution: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{t^2 - 11t + 24}{3t^2 + 12t} \times \dfrac{-6t - 24}{t - 3} $ First factor the quadratic. $p = \dfrac{(t - 3)(t - 8)}{3t^2 + 12t} \times \dfrac{-6t - 24}{t - 3} $ Then factor out any other terms. $p = \dfrac{(t - 3)(t - 8)}{3t(t + 4)} \times \dfrac{-6(t + 4)}{t - 3} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (t - 3)(t - 8) \times -6(t + 4) } { 3t(t + 4) \times (t - 3) } $ $p = \dfrac{ -6(t - 3)(t - 8)(t + 4)}{ 3t(t + 4)(t - 3)} $ Notice that $(t + 4)$ and $(t - 3)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ -6\cancel{(t - 3)}(t - 8)(t + 4)}{ 3t(t + 4)\cancel{(t - 3)}} $ We are dividing by $t - 3$ , so $t - 3 \neq 0$ Therefore, $t \neq 3$ $p = \dfrac{ -6\cancel{(t - 3)}(t - 8)\cancel{(t + 4)}}{ 3t\cancel{(t + 4)}\cancel{(t - 3)}} $ We are dividing by $t + 4$ , so $t + 4 \neq 0$ Therefore, $t \neq -4$ $p = \dfrac{-6(t - 8)}{3t} $ $p = \dfrac{-2(t - 8)}{t} ; \space t \neq 3 ; \space t \neq -4 $